A complex affine variety $X$ of dimension at least $1$ is never compact in the classical topology.
This question is exercise 2.36b, on page 20, in Gathmann's notes has already been asked before (here and here), but the methods used there are not really in line with the rest of Gathmann's text. I am looking for an answer, that is coherent with the rest of text.
My idea so far is the following: We note that:
Lemma 1: Let $f$ be a complex polynomial of degree $n \ge 2$. Then $\lvert V(f) \rvert = \infty$.
Proof: Fix arbitrary $a_2,\ldots,a_n \in \mathbb{C}$. By the fundamental theorem of algebra we know that $f$ splits into linear factors, and by this decomposition we see that $f(x_1,a_2,\ldots,a_n)$ has as a root.
On the other hand we note that
Lemma 2: Let $f$ be a complex polynomial of degree $n = 1$. Then $\lvert V(f) \rvert = 1$.
From the proof Lemma 1 we can further see that
Lemma 3: Let $f$ be a complex polynomial of degree $n \ge 2$. Then $V(f)$ is unbounded.
However, I fail to see how to use that $X$ has dimension $\ge 1$ to apply these observations. (The abstract definition via ascending chains is a bit impractical.) Could you please give me a hint?
Edit 1: I am looking for a proof that is less heavy in terms of the machinery used than the ones in the questions linked above.
Edit 2: In an old version of Gathmann's notes (p.49)the exercise is stated a bit differently:
I understand that the Zariski topology on $\mathbb{A}^1$ equals the cofinite topology. But I do not see how to go on from here.
I don't know if the lemmas you give are usefull for this problem (unless $X$ is a hypersurface, in which case, your problem is a direct conequence of lemma 3). However, if you want a proof, you know by Noether normalisation that there exists a finite injective morphism $\iota : \mathbb{C}[X_1,\ldots,X_d] \hookrightarrow \Gamma(X,\mathbb{A}^1)$. Let $f : X \rightarrow \mathbb{A}^d$ be the morphism such that $\iota = f^*$.
$\iota$ is injective thus $f$ is dominant. It implies that $f(X)$ has a non-empty Zariski interior, which implies that it is dense for the Euclidean topology. In particular, $X$ can't be compact, or else $f(X)$ would be, and it wouldn't be dense.